The residue theorem can be used to compute integrals of functions that do not have an obvious primitive. For this, we first need to find a suitable holomorphic function and contour to which we can apply the residue theorem, which is guesswork. The original integral is often a part of this contour and one shows that the rest of contour integral tends to zero.
We give three examples that illustrate the general technique.
Example 11.1
Let \(f(z)=\frac{1}{1+z^4},\) which is holomorphic except for poles of order one at all points \(\ze^{1+2k},\) \(k=0,1,2,3,\) where \(\ze=e^{i\pi/4}.\) For \(R>0\) introduce the semi-circular contour \(\eta_R\colon[0,\pi]\to\C,\) \(\eta_R(t)=Re^{it}\) and let \(\ga_R\) be the concatenation of \(\eta_R\) with the real interval \([-R,R],\) parametrized as in Example 6.1.
The residues are \[\Res_{\ze^{1+2k}}(f)=\frac{-\ze^{1+2k}}{4}.\]
When \(R>1,\) the only isolated singularities inside the contour are \(\ze, \ze^3,\) hence \(W_\ze(\ga_R)=W_{\ze^3}(\ga_R)=1\) (more formally, this holds by Example 10.3 and Proposition 10.2) and \(W_{\ze^5}(\ga_R)=W_{\ze^7}(\ga_R)=0\) (more formally, by Cauchy’s Theorem 7.1). By the residue Theorem 10.1, \[\int_{\ga_R}f(z)dz=2\pi i\bigl[W_\ze(\ga_R)\Res_\ze(f)+W_{\ze^3}(\ga_R)\Res_{\ze^3}(f)\bigr]=\frac{\pi}{\sqrt{2}}\]
The contour integral decomposes as \[\frac{\pi}{\sqrt{2}}=\int_{\ga_R}f(z)dz=\int_{\eta_R}f(z)dz+\int_{[-R,R]}f(z)dz\]
and we have \[\int_{-\iy}^{+\iy}f(z)dz=\lim_{R\to+\iy}\int_{[-R,R]}f(z)dz.\]
We claim that \(\int_{\eta_R}f(z)dz\to0\) as \(R\to+\iy.\) We have \(L(\eta_R)=\pi R\) and \[|f(z)|=\frac{1}{|z^4+1|}\overset{}{\leqslant}\frac{1}{R^4-1}\] where the inequality follows by Equation 1.7.
Hence Equation 6.7 implies \(\left|\int_{\eta_R}f(z)dz\right|\leqslant\frac{\pi R}{R^4-1},\) which indeed tends to zero as \(R\to+\iy.\) Putting things together, we get \[\int_{-\iy}^{+\iy}\frac{dx}{1+x^4}=\frac{\pi}{\sqrt{2}}.\]
More generally, we can use the approach above to compute all integrals \(\int_{-\iy}^{+\iy}f(z)dz\) of rational functions \(f(z)=\frac{p(z)}{q(z)}\) with \(\deg(q)\geqslant\deg(p)+2\) and \(q(x)\neq0\) for all \(x\in\R.\)
Integrals made only of trigonometric functions can often be evaluated using the residue theorem and the following method.
Example 11.2
If \(x\in\R,\) then \(\cos(x)=\frac{z+1/z}{2},\) \(\sin(x)=\frac{z-1/z}{2i}\) for \(z=e^{ix}.\) Using this, we can change variables and use \(dz=izdx\) to rewrite \[\int_0^{2\pi}\frac{\cos(x)^2}{\sin(x)+2}dx=\int_{\6 D_1(0)}\frac{(1+z^2)^2}{2z^2(z^2+4iz-1)}dz\]
as a contour integral, which is evaluated using the residue theorem. The singularities of \(f(z)=\frac{(1+z^2)^2}{2z^2(z^2+4iz-1)}=\frac{(1+z^2)^2}{2z^2(z-z_+)(z-z_-)}\) are at \(z=0\) (pole of order two) and \(z_\pm=-i(2\pm\sqrt{3})\) (poles of order one). Using Proposition 10.1(c), we find that the residues are \[\begin{align*}
\Res_0(f)&=\lim_{z\to0}\frac{d}{dz}\frac{(1+z^2)^2}{2(z^2+4iz-1)}=-2i,\\
\Res_{z_-}(f)&=\lim_{z\to z_-}\frac{(1+z^2)^2}{2z^2(z-z_+)}=\frac{(1+z_-^2)^2}{2z_-^2(z_--z_+)}=i\sqrt{3}.
\end{align*}\] Now the residue theorem implies \[\int_{\6 D_1(0)}\frac{(1+z^2)^2}{2z^2(z^2+4iz-1)}dz=2\pi i(-2i+i\sqrt{3})=2\pi(\sqrt{3}-2).\]
This example can be generalized to compute integrals over \([0,2\pi]\) of rational algebraic functions of \(\sin(x)\) and \(\cos(x).\)
The following integral was already computed in calculus using polar coordinates in the plane.
Example 11.3
\(\int_{-\iy}^{+\iy} e^{-x^2}dx=\sqrt{\pi}\)
Let \(a=(1+i)\sqrt{\frac{\pi}{2}}.\) Then \(a^2=i\pi.\) The function \(f(z)=\frac{e^{-z^2}}{1+e^{-2az}}\) is holomorphic except for poles of order one at all points \(a\left(k+\frac12\right),\) \(k\in\Z\) (simple zeros of the denominator). Using the power series expansion of the exponential, we find \[\begin{align*}
1+e^{-2az}&=1+e^{-2a\left(z-\frac{a}{2}\right)}e^{a^2}=1-e^{-2a\left(z-\frac{a}{2}\right)}\\
&=2a\left(z-\frac{a}{2}\right)-\frac{1}{2!}4a^2\left(z-\frac{a}{2}\right)^2+\ldots
\end{align*}\] and this gives the residue \[\Res_{a/2}(f) = \lim_{z\to a/2}\frac{e^{-z^2}(z-\frac{a}{2})}{1+e^{-2az}}=\frac{e^{-a^2/4}}{2a}=\frac{-i}{2\sqrt{\pi}}.\]
From \(a^2=i\pi\) it is not hard to check \[f(z)-f(z+a)=e^{-z^2}.\]
Hence the integral over \(z\in[-\iy,+\iy]\) can be computed using the two horizontal parts of the following contour \(\ga_R\) as \(R\to+\iy.\)
The only singularity inside the contour is \(a/2,\) so the residue theorem gives \[\int_{\ga_R} f(z)dz=2\pi i\frac{-i}{2\sqrt{\pi}}=\sqrt{\pi}.\]
The contour integral over \(\ga_R\) splits into four parts. We estimate the vertical parts \(\eta_{\pm R}\) of the integral as follows. Let \(z=x+iy\) with \(x=\pm R\) and \(y\in[0,\sqrt{\pi/2}].\) Then \[|f(z)|\leqslant\frac{e^{-\Re(z^2)}}{1-e^{-\Re(2 az)}}=\frac{e^{y^2-x^2}}{1-e^{\sqrt{2\pi}(y-x)}}\leqslant\frac{e^{\pi/2-x^2}}{1-e^{-\sqrt{2\pi}x}}\]
tends to zero as \(|x|\to+\iy,\) uniformly in \(y\in[0,\sqrt{\pi/2}].\) Hence Equation 6.7 implies \[\int_{\eta_{\pm R}} f(z)dz\xrightarrow{R\to\iy}0\]
and we conclude \[\begin{align*}
\sqrt{\pi}&=\int_{\ga_R} f(z)dz\\
&=\int_{-R}^{+R} f(z)dz+\int_{R}^{-R} f(z+a)dz+\int_{\eta_{R}} f(z)dz+\int_{\eta_{-R}} f(z)dz\\
&\xrightarrow{R\to+\iy}\int_{-\iy}^{+\iy} f(z)dz-\int_{-\iy}^{+\iy} f(z+a)dz=\int_{-\iy}^{+\iy} e^{-x^2}dx.
\end{align*}\]
References
Dowson, H. R. 1979.
“Serge Lang, Complex Analysis (Addison-Wesley,
1977), Xi + 321 Pp., £1200.” Proceedings of the Edinburgh
Mathematical Society 22 (1): 65–65.
https://doi.org/10.1017/S0013091500027838.
Freitag, E., and R. Busam. 2009.
Complex Analysis.
Universitext. Springer Berlin Heidelberg.
https://books.google.co.uk/books?id=3xBpS-ZKlgsC.
Howie, J. M. 2012.
Complex Analysis. Springer Undergraduate
Mathematics Series. Springer London.
https://books.google.co.uk/books?id=0FZDBAAAQBAJ.
Jameson, G. J. O. 1987.
“H. A. Priestley, Introduction to Complex
Analysis (Oxford University Press, 1985), 197 Pp., £8.50.”
Proceedings of the Edinburgh Mathematical Society 30 (2):
325–26.
https://doi.org/10.1017/S0013091500028406.
Remmert, R. 1991.
Theory of Complex Functions. Graduate Texts
in Mathematics. Springer New York.
https://books.google.co.uk/books?id=uP8SF4jf7GEC.
Thomson, B. S., J. B. Bruckner, and A. M. Bruckner. 2001.
Elementary
Real Analysis. Prentice-Hall.
https://books.google.co.uk/books?id=6l_E9OTFaK0C.
